Thread: arithmetic progression

A training programme requires a footballer to run circuits of a football field every day . On each day , he runs three circuit more than he did the day before . On the fifth day he runs 14 circuits . Calculate the number of circuits covered by him altogether from the end of the n th day to the end of the ( 2n ) th dat , in terms of n


Can someone pls explain to me .Thank s

Term n of an arithmetic series given by
Tn= a + (n-1)d

where a = first term and d = common difference
d is given = 3
T5 = a + (5-1)3 = a + 12 = 14
so a = 2

Sn the summation of the arithmetic series is given by

Sn= n/2[2a +(n-1)d]

so in this case Sn= n/2[4+3(n-1)]

and S2n= 2n/2[4+3(2n-1)]

S2n - Sn reduces to n/2(9n+1)